As the electron in a Bohr orbit of a Hydrogen | vedclass

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(A) The total energy of an electron in the $n$-th orbit is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For $n = 2$,$E_2 = -\frac{13.6}{4} = -3.4 	e

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$K$ becomes $4$ times,$U$ becomes $4$ times

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(A) The total energy of an electron in the $n$-th orbit is given by $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.For $n = 2$,$E_2 = -\frac{13.6}{4} = -3.4 	ext{ eV}$.For $n = 1$,$E_1 = -\frac{13.6}{1} = -13.6 	ext{ eV}$.We know that kinetic energy $K = -E$ and potential energy $U = 2E$.At $n = 2$: $K_2 = 3.4 	ext{ eV}$ and $U_2 = -6.8 	ext{ eV}$.At $n = 1$: $K_1 = 13.6 	ext{ eV}$ and $U_1 = -27.2 	ext{ eV}$.Ratio of kinetic energies: $\frac{K_1}{K_2} = \frac{13.6}{3.4} = 4$.Ratio of potential energies: $\frac{U_1}{U_2} = \frac{-27.2}{-6.8} = 4$.Thus,both kinetic energy and potential energy increase by a factor of $4$.

As the electron in a Bohr orbit of a Hydrogen atom transitions from state $n = 2$ to $n = 1$,how do the kinetic energy $K$ and potential energy $U$ change?

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